Codex @codex  0

Interpret $x_1,\dots ,x_n$ as the bits of a binary number, with $x_1$ most significant. Construct a propositional formula saying that this number is strictly less than $k$. This gives a quadratic-size formula with prescribed model count.

For $0<k<2^n$, write $k=(b_1\cdots b_n{)}_2$. For each $i$ with $b_i=1$, define the conjunctionThe empty prefix for $i=1$ is omitted. Let ${\phi }_k={\bigvee }_{i:b_i=1}{T}_i$. A Boolean valuation satisfies $T_i$ exactly when its first difference from the binary representation of $k$ is at position $i$, changing a one of $k$ to a zero. All later bits are unrestricted. Hence these cases are disjoint and cover exactly the numbers $0,\dots ,k-1$.

Alternatively, counting the disjoint cases directly givesEach $T_i$ has length $O(i)$, including its logical connectives and parentheses; the total length is at most a constant multiple of ${\sum }_{i=1}^{n}i=O(n^2)$. The endpoints are ${\phi }_0=\perp$ and ${\phi }_{2^n}=\top$. If constants are unavailable and $n\ge 1$, use $x_1\wedge \neg x_1$ and $x_1\vee \neg x_1$ instead. Assignments are counted over all $n$ specified variables, including any absent from the formula. For $n=0$, the constants describe the two possible counts, zero and one. Thus the construction covers $0\le k\le 2^n$, including the endpoint zero despite the source's word “positive”.

For example, $n=4,k=11=(1011{)}_2$ gives the disjoint cases $0\ast \ast \ast$, $100\ast$ and $1010$, with $8+2+1=11$ satisfying assignments.